College Trigonometry, 2011a

748 Foundations of Trigonometry
y
1
y
1
π
3
π
π
3
1
x
π
3
1
x
3. From the table of common values, we see that π 4
has a cotangent of 1, which means the
solutions to cot(θ) =−1 have a reference angle of π 4
. To find the quadrants in which our
solutions lie, we note that cot(θ) = x y
for a point (x, y) on the Unit Circle where y ≠0. If
cot(θ) is negative, then x and y must have different signs (i.e., one positive and one negative.)
Hence, our solutions lie in Quadrants II and IV. Our Quadrant II solution is θ = 3π 4
+2πk,
and for Quadrant IV, we get θ = 7π 4
+2πk for integers k. Can these lists be combined? Indeed
they can - one such way to capture all the solutions is: θ = 3π 4
+ πk for integers k.
y
y
1
1
π
4
π
4
1
x
π
π
4
1
x
We have already seen the importance of identities in trigonometry. Our next task is to use use the
Reciprocal and Quotient Identities found in Theorem 10.6 coupled with the Pythagorean Identity
found in Theorem 10.1 to derive new Pythagorean-like identities for the remaining four circular
functions. Assuming cos(θ) ≠ 0, we may start with cos 2 (θ) +sin 2 (θ) = 1 and divide both sides
by cos 2 (θ) to obtain 1 + sin2 (θ)
cos 2 (θ) = 1 . Using properties of exponents along with the Reciprocal
cos 2 (θ)
and Quotient Identities, this reduces to 1 + tan 2 (θ) = sec 2 (θ). If sin(θ) ≠ 0, we can divide both
sides of the identity cos 2 (θ)+sin 2 (θ) =1bysin 2 (θ), apply Theorem 10.6 once again, and obtain
cot 2 (θ) + 1 = csc 2 (θ). These three Pythagorean Identities are worth memorizing and they, along
with some of their other common forms, are summarized in the following theorem.